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- What does $QAQ^{-1}$ actually mean? - Mathematics Stack Exchange
1 $\begingroup$ When one thinks of matrix products like that, it's helpful to remember that matrices, unlike vectors, have two sets of bases: one for the domain and one for the range Thinking of applying a vector on to the right, we get that the transformation "unrotates" the vector, applies the original transformation in the original basis
- arithmetic - Formal proof for $(-1) \times (-1) = 1$ - Mathematics . . .
The Law of Signs $\rm\: (-x)(-y) = xy\:$ isn't normally assumed as an axiom Rather, it is derived as a consequence of more fundamental Ring axioms $ $ [esp the distributive law $\rm\,x(y+z) = xy + xz\,$], laws which abstract the common algebraic structure shared by familiar number systems
- Why is $1^{\\infty}$ considered to be an indeterminate form
The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form
- Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
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- Why is $1 i$ equal to $-i$? - Mathematics Stack Exchange
There are multiple ways of writing out a given complex number, or a number in general Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example
- abstract algebra - Prove that 1+1=2 - Mathematics Stack Exchange
The main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps The work of G Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do
- notation - What does this symbol mean? (looks like a 1 with double . . .
$\begingroup$ For those wondering, the symbol (𝟙 in Unicode; fonts may lack it) is the blackboard bold open face double-struck digit one — the same style as ℕ, ℝ, ℤ, etc $\endgroup$ –
- If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?
A-1 A means that first we apply A transformation then we apply A-1 transformation When we apply A transformation we reach some plane having some different basis vectors but after apply A-1 we again reach to the plane have basis i ^ (0,1) and j ^ (1,0) It means that after applying A-1 we reach to the transformation which does nothing
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